Using the formula $$v = \sqrt{\dfrac{\gamma P}{\rho }}ργP$$​​  explain why the speed of sound in air

(a) does not depend upon pressure.

(b) increases with temperature and humidity.

(c) increases with humidity.

Asked by Pragya Singh | 1 year ago |  120

##### Solution :-

Given,

v = $$\sqrt{\dfrac{\gamma P}{\rho }}ργP​​$$

We know,

PV = nRT ( for n moles of ideal gas )

= PV = RT ($$\dfrac{m}{M}$$)

Where, m is the total mass and M is molecular mass of the gas.

therefore, P

= m ($$\dfrac{RT}{M}$$)

= P = $$\dfrac{\rho RT}{M}MρRT​$$

$$\dfrac{P}{\rho} = \dfrac{ RT}{M}​=MRT​$$

(a) For a gas at a constant temperature,

$$\dfrac{P}{\rho}$$ = constant

Thus, as P increases ρ and vice versa.

This means that $$\dfrac{P}{\rho}$$ ratio always remains constant meaning

v =$$\sqrt{\dfrac{\gamma P}{\rho }}ργP​​$$
constant i.e., velocity of sound does not depend upon the pressure of the gas.

( b ) Since, $$\dfrac{P}{\rho}$$

$$= \dfrac{ RT}{M}ρP​=MRT​$$

v =  $$\sqrt{\dfrac{\gamma P}{\rho }}ργP​​$$

v =$$\sqrt{\dfrac{\gamma RT}{M }}MγRT​​$$

We can see that v $$\propto \sqrt{T}∝T$$​ i.e., speed of sound increases with temperature.

(c) When humidity increases, effective density of the air decrease.

This means v $$\propto \dfrac{1}{\sqrt{\rho }}v∝ρ​1$$​, thus velocity increases.

Answered by Abhisek | 1 year ago

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